Number of ways to appoint a class leaderships with 36 students

A class has 16 girls and 20 boys. The class leadership should be made by 4 students that should has at least one girl. In how many ways the choices can be made.

I think this can be solve with combinations, and every group of class leadership is divided in four possible cases.

1. GIRL GIRL GIRL GIRL
2. BOY GIRL GIRL GIRL
3. BOY BOY GIRL GIRL
4. BOY BOY BOY GIRL

What to do next?

Method N°1

As you said, there are 4 possible ways to have a class leadership

1. 4 Girls
2. 3 Girls 1 Boy
3. 2 Girls 2 Boys
4. 1 Girl 3 Boys

From there, we simply need to compute each of these cases and to sum these up:

1. Case 1 gives $\binom{16}{4}$ because we have $\binom{16}{4}$ ways to pick 4 girls from a group of 16.

2. Case 2 gives $\binom{16}{3} \times \binom{20}{1}$ because we have $\binom{16}{3}$ ways to pick 3 people in a group of 16 and $\binom{20}{1}$ ways to pick 1 people in a group of twenty.

3. Case 3 gives $\binom{16}{2} \times \binom{20}{2}$ (We repeat the same process here, and for the 4th case)

4. Case 4 gives $\binom{16}{1} \times \binom{20}{3}$

So the total number of possible class leaderships are $$\binom{16}{4} + \binom{16}{3} \times \binom{20}{1} + \binom{16}{2} \times \binom{20}{2} + \binom{16}{1} \times \binom{20}{3} = 54 \ 060$$

Method N°2

We can also simply notice that the question asks for the number of groups of four people, except the groups that are only made of boys.

This lead to a simple computation: the number of ways to choose 4 people among 36, minus the number of ways to choose 4 boys among 20.

Thus, we have:

$$\binom{36}{4} - \binom{20}{4} = 54 \ 060$$

Which gives the same result as with the first method